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Chemistry · chapter PYQs

Coordination Compounds, on the exam screen.

12 questions from NEET 2026, Re-NEET 2026, NEET 2020, 12 minutes, marked +4 / −1. See them with solutions.

Coordination Compounds · PYQs

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  1. Question 1 · NEET 2026 · Q53 · +4 / −1

    Match List I (Complex) with List II (Type of isomerism):

    List I

    List II

    a) [Pt(NH3)2Cl2][\mathrm{Pt}(\mathrm{NH}_{3})_{2}\mathrm{Cl}_{2}]

    i) Optical

    b) [Co(en)3]3+[\mathrm{Co}(\mathrm{en})_{3}]^{3+}

    ii) Solvate

    c) [Co(NH3)5NO2]Cl2[\mathrm{Co}(\mathrm{NH}_{3})_{5}\mathrm{NO}_{2}]\mathrm{Cl}_{2}

    iii) Geometrical

    d) [Cr(H2O)6]Cl3[\mathrm{Cr}(\mathrm{H}_{2}\mathrm{O})_{6}]\mathrm{Cl}_{3}

    iv) Linkage

    Answer and explanation

    Answer: (4)

    [Pt(NH3)2Cl2][\mathrm{Pt}(\mathrm{NH}_{3})_{2}\mathrm{Cl}_{2}] shows cis/trans (geometrical) isomerism (III); [Co(en)3]3+[\mathrm{Co}(\mathrm{en})_{3}]^{3+} is a tris-chelate that is chiral, showing optical isomerism (I); [Co(NH3)5NO2]Cl2[\mathrm{Co}(\mathrm{NH}_{3})_{5}\mathrm{NO}_{2}]\mathrm{Cl}_{2} shows linkage isomerism through the ambidentate NO2−\mathrm{NO}_{2}^{-}/ONO−\mathrm{ONO}^{-} ligand (IV); and [Cr(H2O)6]Cl3[\mathrm{Cr}(\mathrm{H}_{2}\mathrm{O})_{6}]\mathrm{Cl}_{3} shows solvate (hydrate) isomerism (II). Hence A-III, B-I, C-IV, D-II.

    figure
  2. Question 2 · NEET 2026 · Q63 · +4 / −1

    Which one of the following is an ambidentate ligand?

    Answer and explanation

    Answer: (3)

    An ambidentate ligand has two different donor atoms and can bind through either one. Thiocyanate (SCN-) can coordinate through sulphur (-SCN) or nitrogen (-NCS), so it is ambidentate. Ethane-1,2-diamine and oxalate are bidentate chelating ligands and ethylenediaminetetraacetate ion (EDTA) is hexadentate, but each of their donor sets is fixed, so they are not ambidentate.

  3. Question 3 · NEET 2026 · Q68 · +4 / −1

    Match List I (Complex/ion) with List II (Shape/geometry) and choose the correct answer.

    List I

    List II

    P) [Pt(Cl2)(NH3)2][Pt(Cl_2)(NH_3)_2]

    I) Octahedral

    Q) [Co(NH3)6]Cl3[Co(NH_3)_6]Cl_3

    II) Trigonal bipyramidal

    R) [NiCl4]2−[NiCl_4]^{2-}

    III) Square planar

    S) [Fe(CO)5][Fe(CO)_5]

    IV) Tetrahedral

    Answer and explanation

    Answer: (2)

    [PtCl2_{2}(NH3_{3})2_{2}] uses dsp2^{2} hybridisation at Pt2+^{2+} → square planar (III); [Co(NH3_{3})6_{6}]3+^{3+} is d2^{2}sp3^{3} → octahedral (I); [NiCl4_{4}]2−^{2-} has sp3^{3} Ni2+^{2+} with weak-field Cl−^- → tetrahedral (IV); [Fe(CO)5_{5}] is sp3^{3}d/dsp3^{3} → trigonal bipyramidal (II). This gives P-III, Q-I, R-IV, S-II.

  4. Question 4 · Re-NEET 2026 · Q47 · +4 / −1

    Given below are two statements. (Given : oxH2=HOOC−COOH\mathrm{oxH}_{2} = \mathrm{HOOC}-\mathrm{COOH})

    Statement I: [Fe(ox)3]3−[\mathrm{Fe}(\mathrm{ox})_{3}]^{3-} is chiral.
    Statement II: trans-[Cr(H2O)2(ox)2]−[\mathrm{Cr}(\mathrm{H}_{2}\mathrm{O})_{2}(\mathrm{ox})_{2}]^- is chiral.

    In light of the above statements, choose the most appropriate answer from the options given below:

    Answer and explanation

    Answer: (3)

    A tris-chelate of type [M(AA)3][\mathrm{M}(\mathrm{AA})_{3}] has no plane or centre of symmetry and exists as the non-superimposable delta and lambda propeller forms, so [Fe(ox)3]3−[\mathrm{Fe}(\mathrm{ox})_{3}]^{3-} is indeed chiral. For the bis-chelate [Cr(H2O)2(ox)2]−[\mathrm{Cr}(\mathrm{H}_{2}\mathrm{O})_{2}(\mathrm{ox})_{2}]^-, only the CIS isomer is chiral; the TRANS isomer, with the two aqua ligands on opposite sides, possesses a plane of symmetry and is therefore achiral. Statement-II names the wrong isomer. NCERT Class 12, Coordination Compounds (isomerism).

  5. Question 5 · Re-NEET 2026 · Q58 · +4 / −1

    According to crystal field theory, the correct order of ligands with respect to their decreasing order of field strength is

    Answer and explanation

    Answer: (1)

    The spectrochemical series orders ligands by the size of the crystal field splitting they produce. Carbon monoxide is the strongest field ligand in this set because it accepts electron density into its π∗\pi^{*} orbitals; ammonia follows as a stronger sigma donor than water; and chloride, being a weak-field pi donor, is the weakest. The accepted sequence is CO>CN−>NH3>H2O>F−>Cl−\mathrm{CO} > \mathrm{CN}^- > \mathrm{NH}_{3} > \mathrm{H}_{2}\mathrm{O} > \mathrm{F}^- > \mathrm{Cl}^-. NCERT Class 12, Coordination Compounds (crystal field theory).

  6. Question 6 · Re-NEET 2026 · Q66 · +4 / −1

    Among the species given below, the spin-only magnetic moment is highest for (Given: Atomic number of Ti = 22, Mn = 25, Fe = 26 and Co = 27)

    Answer and explanation

    Answer: (1)

    All four are octahedral with strong-field or moderate-field ligands, so the electrons pair up in t2gt_{2g} before entering ege_g. Counting d electrons: Mn3+\mathrm{Mn}^{3+} is d4d^4, giving low-spin t2g4t_{2g}^4 with TWO unpaired electrons; Fe3+\mathrm{Fe}^{3+} is d5d^5, giving t2g5t_{2g}^5 with one; Co3+\mathrm{Co}^{3+} is d6d^6, giving t2g6t_{2g}^6 with none; Ti3+\mathrm{Ti}^{3+} is d1d^1 with one. Since μ=n(n+2)\mu = \sqrt{n(n+2)} rises with n, the manganese complex has the highest moment at about 2.83 Bohr magneton (BM). NCERT Class 12, Coordination Compounds.

  7. Question 7 · Re-NEET 2026 · Q68 · +4 / −1

    The formula of tetraammineaquachloridocobalt(III) chloride is

    Answer and explanation

    Answer: (4)

    The name lists the coordination sphere first: four ammine, one aqua and one chlorido ligand around cobalt(III). Charge balance inside the brackets gives +3+0+0−1=+2+3 + 0 + 0 - 1 = +2, so two chloride counter-ions are needed outside to neutralise the complex cation. The word 'chloride' at the end names those counter-ions, not a ligand. NCERT Class 12, Coordination Compounds (IUPAC nomenclature).

  8. Question 8 · Re-NEET 2026 · Q78 · +4 / −1

    The complex which has facial and meridional isomers is (Given : py = pyridine and en = H2N−CH2−CH2−NH2\mathrm{H}_{2}\mathrm{N}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{NH}_{2})

    Answer and explanation

    Answer: (1)

    Facial and meridional isomerism is specific to octahedral complexes of the type [Ma3b3][\mathrm{Ma}_{3}\mathrm{b}_{3}]. In the fac isomer the three identical ligands occupy one triangular face; in the mer isomer they lie in a plane through the metal. [Cr(py)3Cl3][\mathrm{Cr}(\mathrm{py})_{3}\mathrm{Cl}_{3}] has exactly this 3:3 arrangement. NCERT Class 12, Coordination Compounds (geometrical isomerism).

  9. Question 9 · Re-NEET 2026 · Q87 · +4 / −1

    A 1 : 3 electrolyte in an aqueous solution is

    Answer and explanation

    Answer: (3)

    Only the ions OUTSIDE the coordination sphere are released in solution; ligands bound to the metal stay put. [Co(NH3)6]Cl3[\mathrm{Co}(\mathrm{NH}_{3})_{6}]\mathrm{Cl}_{3} has all six ammonia molecules coordinated and three chlorides as counter-ions, so it dissociates into one complex cation and three chloride ions - a 1 : 3 electrolyte. NCERT Class 12, Coordination Compounds (Werner's theory, primary and secondary valence).

  10. Question 10 · NEET 2020 · Q98 · +4 / −1

    Which of the following is the correct order of increasing field strength of ligands to form coordination compounds?

    Answer and explanation

    Answer: (4)

    In the spectrochemical series relevant here, ligand field strength rises as SCN−\mathrm{SCN}^{-} < F−\mathrm{F}^{-} < C2O42−\mathrm{C}_{2}\mathrm{O}_{4}^{2-} < CN−\mathrm{CN}^{-}. Chapter: Coordination Compounds.

  11. Question 11 · NEET 2020 · Q99 · +4 / −1

    The calculated spin-only magnetic moment of Cr2+^{2+} ion is

    Answer and explanation

    Answer: (1)

    The spin-only magnetic moment μ\mu is given by μ=n(n+2) BM\mu = \sqrt{n(n+2)}\ BM, where n is the number of unpaired electrons. For Cr2+\mathrm{Cr}^{2+}, the electronic configuration is 3d43d^4, with 4 unpaired electrons. Substituting n = 4, μ=4(4+2)=24≈4.90 BM\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90\ BM. Chapter: Coordination Compounds.

  12. Question 12 · NEET 2020 · Q122 · +4 / −1

    Urea reacts with water to form A, which decomposes to form B. When B is passed through Cu2+^{2+}(aq), a deep-blue solution C is formed. What is the formula of C?

    Answer and explanation

    Answer: (1)

    Urea hydrolysis ultimately releases NH3_{3}. Excess ammonia coordinates Cu2+^{2+} to form the deep-blue tetraamminecopper(II) ion [Cu(NH3_{3})4_{4}]2+^{2+}. Chapter: Coordination Compounds.